The problem statement is as follows -
Consider the parametrised surface $$ x(u,v) = \bigg(\sin u\cos v , \sin u\sin v , \cos u + \log(\tan\frac{u}{2}) +\phi(v) \bigg) $$
where $\ \phi\ $ is a differentiable function. Prove that
a. The curves $\ v \ $ = const. are contained in planes which pass through the z axis and intersect the surface under a constant angle $\ \theta\ $ given by
$$ \cos \theta = \frac{\phi'}{\sqrt{1+(\phi')^2}} $$
My attempt :
For given parametrized surface $$ x(u,v) = \bigg(\sin u\cos v , \sin u\sin v , \cos u + \log(\tan\frac{u}{2}) +\phi(v) \bigg) $$ with $ \ \phi \ $ - differentiable function we can write
$ x_u = \bigg( \cos u\cos v \ ,\cos u \sin v \ , \cot u \cos u \bigg)$
$ x_v = \bigg( -\sin u\sin v \ ,\sin u \cos v \ , \phi' \bigg)$
$x_u\wedge x_v = \begin{vmatrix}
i & j & k\\
\cos u \cos v & \cos u \sin v & \cot u \cos u \\
-\sin u \sin v & \sin u \cos v & \phi'
\end{vmatrix} \\ = i\bigg( \phi' \cos u \sin v - \cos ^2 u \cos v \bigg) - j\bigg( \phi' \cos u \cos v + \cos ^2 u \sin v \bigg) + k\bigg( \sin u \cos u \bigg) $
The unit normal N = $\frac{x_u\wedge x_v}{||x_u\wedge x_v||}$ .
We first calculate $||x_u\wedge x_v||$
$$||x_u\wedge x_v|| \ = \ \sqrt{(\phi')^2\cos ^2u \sin ^2 v + \cos ^4u \cos ^2v - 2\phi'\cos ^3u \sin v \cos v + (\phi')^2\cos ^2u \cos ^2v \\ + \cos ^4u \sin ^2v + 2\phi'\cos ^3u \cos v\sin v + \sin ^2u \cos ^2u }$$ $||x_u\wedge x_v|| = \cos u \sqrt{1 + (\phi')^2}$
Hence, the unit normal to surface $ x(u,v) $ is given by
$$N = \frac{1}{\sqrt{1 + (\phi')^2}} \bigg( \phi' \sin v - \cos u \cos v , \ - \phi' \cos v - \cos u \sin v , \ \sin u \bigg)$$
Now, the unit normal vector to the plane passing through the curve $ v $ = const. and the z-axis , say V, will be given by
$ V = (-\sin u, \cos u , 0 ) $ and then $ \theta $ will be
$ \cos\theta = N\cdot V$
$$ \cos \theta = \frac{\phi' (-\sin ^2 v ) + \cos u \cos v \sin v - \phi' \cos ^2 v - \cos u \sin v \cos v }{\sqrt{1+ (\phi')^2}}$$
$$ \cos \theta = \frac{ - \phi' } {\sqrt{1+ (\phi')^2}}$$
Why am I getting the undesired -ve sign here? Is it due to the wrong direction of vector V or something else went wrong?
